# RS Aggarwal Class 9 Solutions Chapter 10 - Area Ex 10A (10.1)

## Exercise 10.1: Area

Q.1: In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P.

Prove that:Â ar.( â–³ABP$\triangle ABP$ ) = ar. (quad. ABCD)

Sol:

Given: ABCD is a quadrilateral in which through D, a line is drawn parallel to AC which meets BC produced in P.

To prove: ar(Î”ABP)=ar(quad.ABCD)$ar\left ( \Delta ABP \right )=ar\left ( quad.ABCD \right )$

Proof: Î”ACPandÎ”ACD$\Delta ACP \:and\:\Delta ACD$ have same base AC and lie between parallel lines AC and DP.

Therefore, ar(Î”ACP)=ar(Î”ACD)$ar\left ( \Delta ACP \right )=ar\left ( \Delta ACD \right )$

Adding ar(Î”ACP)$ar\left ( \Delta ACP \right )$ on both sides, we get;

ar(Î”ACP)+ar(Î”ABC)=ar(Î”ACD)+ar(Î”ABC)$ar\left ( \Delta ACP \right )+ar\left ( \Delta ABC \right )= ar\left ( \Delta ACD \right )+ar\left ( \Delta ABC \right )$

Therefore, ar(Î”ABP)=ar(quad.ABCD)$ar\left ( \Delta ABP \right )=ar\left ( quad.ABCD \right )$

Q.2: In the adjoining figure, AD and BC are 2 diagonals of the parallelogram ABCD. The point at which the diagonals intersect is O. Prove that OA = OD.

Sol:

Given: Two triangles, i.e. Î”ABCandÎ”DBC$\Delta ABC\:and\:\Delta DBC$ which have same base BC and points A and D lie on opposite sides of BC and ar(Î”ABC)=ar(Î”DBC)$ar\left ( \Delta ABC \right )=ar\left ( \Delta DBC \right )$

To prove: Â OA = OD

Construction: Draw APâŠ¥BCandDQâŠ¥BC$AP\perp BC\:and\:DQ\perp BC$

Proof: We have:

ar(Î”ABC)=12Ã—BCÃ—AP$ar\left ( \Delta ABC \right )=\frac{1}{2}\times BC\times AP$ and

ar(Î”BCD)=12Ã—BCÃ—DQ$ar\left ( \Delta BCD \right )=\frac{1}{2}\times BC\times DQ$

So, 1/2 x BC x AP = 1/2 x BC x DQ [From (1)]

Therefore, AP = DQ . . . . . . . . . (2)

Now, in Î”AOPandÎ”QOD$\Delta AOP\:and\:\Delta QOD$, we have:

âˆ APO=âˆ DQO=90âˆ˜$\angle APO=\angle DQO=90^{\circ}$

And, âˆ APO=âˆ DOQ=90âˆ˜$\angle APO=\angle DOQ=90^{\circ}$Â [vertically opp. Angles]

AP = DQ [From (2)]

Thus, by Angle-Angle-side criterion of congruence, we have:

Therefore, Î”AOPâ‰…Î”QOD$\Delta AOP\cong \Delta QOD$ [AAS]

The corresponding parts of the congruent triangles are equal.

Therefore, OA = OD [C.P.C.T]

Q.3: In the adjoining figure, AD is one of the medians of a â–³ABC$\triangle ABC$ and P is appointed on AD.

Prove that:

(i) ar.( â–³BDP$\triangle BDP$) = ar.( â–³CDP$\triangle CDP$)

(ii)ar.( â–³ABP$\triangle ABP$ = ar.( â–³ACP$\triangle ACP$)

Sol:

Given: A Î”ABC$\Delta ABC$ in which AD is the median and P is a point on AD.

To prove:

(i) ar(Î”BDP)=ar(Î”CDP)$ar\left ( \Delta BDP \right )=ar\left ( \Delta CDP \right )$

(ii) ar(Î”ABP)=ar(Î”APC)$ar\left ( \Delta ABP \right )=ar\left ( \Delta APC \right )$

Proof:

(i) In Î”BPC$\Delta BPC$, PD is the median. Since median of a triangle divides the triangle into two triangles of equal areas.

So, ar(Î”BDP)=ar(Î”CDP)$ar\left ( \Delta BDP \right )=ar\left ( \Delta CDP \right )$ . . . . . . . . (1)

(ii) In Î”ABC$\Delta ABC$, AD is the median

So, ar(Î”ABD)=ar(Î”ADC)$ar\left ( \Delta ABD \right )=ar\left ( \Delta ADC \right )$

But, ar(Î”BDP)=ar(Î”CDP)$ar\left ( \Delta BDP \right )=ar\left ( \Delta CDP \right )$ [From (1)]

Subtracting ar(Î”BPD)$ar\left ( \Delta BPD \right )$ from both the sides of the equation, we have

Therefore, ar(Î”ABD)âˆ’ar(Î”BPD)=ar(Î”ADC)âˆ’ar(Î”BPD)$ar\left ( \Delta ABD \right )-ar\left ( \Delta BPD \right )=ar\left ( \Delta ADC \right )-ar\left ( \Delta BPD \right )$

=ar(Î”BPD)âˆ’ar(Î”CDP)$=ar\left ( \Delta BPD \right )-ar\left ( \Delta CDP \right )$Â [From (1)]

Therefore, ar(Î”ABP)=ar(Î”ACP)$ar\left ( \Delta ABP \right )=ar\left ( \Delta ACP \right )$

Q.4: In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O.

If BO = OD, prove that

ar.( â–³ABC$\triangle ABC$)=ar. (â–³ADC$\triangle ADC$).

Sol:

Given: A quadrilateral ABCD in which diagonals AC and BD intersect at O and BO = OD

To Prove: ar(Î”ABC)=ar(Î”ADC)$ar\left ( \Delta ABC \right )=ar\left ( \Delta ADC \right )$

Proof: Since OB = OD [given]

So, AO is the median of Î”ABD$\Delta ABD$

Therefore, ar(Î”AOD)=ar(Î”AOB)$ar\left ( \Delta AOD \right )=ar\left ( \Delta AOB \right )$ . . . . . . . . (i)

As OC is the median of Î”CBD$\Delta CBD$

Therefore, ar(Î”DOC)=ar(Î”BOC)$ar\left ( \Delta DOC \right )=ar\left ( \Delta BOC \right )$ . . . . . . . . . (ii)

Adding both sides of (i) and (ii), we get:

ar(Î”AOD)+ar(Î”DOC)=ar(Î”AOB)+ar(Î”BOC)$ar\left ( \Delta AOD \right )+ar\left ( \Delta DOC \right )= ar\left ( \Delta AOB \right )+ar\left ( \Delta BOC \right )$

Therefore, ar(Î”ADC)=ar(Î”ABC)$ar\left ( \Delta ADC \right )=ar\left ( \Delta ABC \right )$

Q.5: ABC is a triangle in which D is the midpoint of BC and E is the midpoint of AD.

Prove that:

ar.( â–³BED$\triangle BED$)= 14ar(â–³ABC)$\frac{1}{4}ar(\triangle ABC)$

Sol:

Given: A Î”ABC$\Delta ABC$ in which AD is a median and E is the mid-point of AD

To prove: ar(Î”BED)=14ar(Î”ABC)$ar\left ( \Delta BED \right )=\frac{1}{4}ar\left ( \Delta ABC \right )$

Proof: Since, ar(Î”ABD)=ar(Î”ACD)$ar\left ( \Delta ABD \right )=ar\left ( \Delta ACD \right )$ Â Â [Since,ADisamedian]$\left [Since, AD\:is\:a\:median \right ]$

i.e. ar(Î”ABD)=12ar(Î”ABC)$ar\left ( \Delta ABD \right )=\frac{1}{2}ar\left ( \Delta ABC \right )$ . . . . . . . (1) [Since,ar(Î”ABC)=ar(Î”ABD)+ar(Î”ADC)]$\left [ Since, ar\left ( \Delta ABC \right )=ar\left ( \Delta ABD \right ) +ar\left ( \Delta ADC \right )\right ]$

Now, as BE is the median of Î”ABD$\Delta ABD$

ar(Î”ABE)=ar(Î”BED)$ar\left ( \Delta ABE \right )=ar\left ( \Delta BED \right )$ . . . . . . . . . (2)

Since, ar(Î”ABD)=ar(Î”ABE)+ar(Î”BED)$ar\left ( \Delta ABD \right )=ar\left ( \Delta ABE \right ) +ar\left ( \Delta BED \right )$ . . . . . . . (3)

Therefore,Â ar(Î”BED)=ar(Î”ABE)$ar\left ( \Delta BED \right )=ar\left ( \Delta ABE \right )$â€¦â€¦..[From(2)]

= 12ar(Î”ABD)$\frac{1}{2}ar\left ( \Delta ABD \right )$ Â [From (2) and (3)]

= 12[12ar(Î”ABC)]$\frac{1}{2}\left [ \frac{1}{2}ar\left ( \Delta ABC \right ) \right ]$ Â [From (1)]

= 14ar(Î”ABC)$\frac{1}{4}ar\left ( \Delta ABC \right )$

Q.6: The vertex A of â–³ABC$\triangle ABC$ is joined to a point D on the side BC. The midpoint AD is E.

Prove that:

ar(â–³BEC$\triangle BEC$) = 12ar(â–³ABC)$\frac{1}{2}ar(\triangle ABC)$

Sol:

Given: A Î”ABC$\Delta ABC$ in which E is the mid-point of the line segment AD where D is a point on BC.

To Prove: ar(Î”BEC)=12ar(Î”ABC)$ar\left ( \Delta BEC \right )=\frac{1}{2}ar\left ( \Delta ABC \right )$

Proof: Since, BE is a median of Î”ABD$\Delta ABD$

So, ar(Î”BDE)=ar(Î”ABE)$ar\left ( \Delta BDE \right )=ar\left ( \Delta ABE \right )$

i.e. ar(Î”BDE)=12ar(Î”ABD)$ar\left ( \Delta BDE \right )=\frac{1}{2}ar\left ( \Delta ABD \right )$ . . . . . . . (1)

As, CE is the median of Î”ADC$\Delta ADC$

So, ar(Î”CDE)=12ar(Î”ACD)$ar\left ( \Delta CDE \right )=\frac{1}{2}ar\left ( \Delta ACD \right )$ . . . . . . . (2)

Adding (i) and (ii), we get:

ar(Î”BDE)+ar(Î”CDE)=12ar(Î”ABD)+12ar(Î”ACD)$ar\left ( \Delta BDE \right )+ar\left ( \Delta CDE \right )=\frac{1}{2}ar\left ( \Delta ABD \right )+\frac{1}{2}ar\left ( \Delta ACD \right )$
ar(Î”BDE)=12[ar(Î”ABD)+ar(Î”ACD)]$ar\left ( \Delta BDE \right )=\frac{1}{2}\left [ ar\left ( \Delta ABD \right )+ar\left ( \Delta ACD \right ) \right ]$

= 12ar(Î”ABC)$\frac{1}{2}ar\left ( \Delta ABC \right )$

Q.7: D is the midpoint of side BC of â–³ABC$\triangle ABC$ and E is the midpoint of BD. if O is the midpoint of AE.

Prove that:

ar.( â–³BOE$\triangle BOE$ ) = 18ar(â–³ABC)$\frac{1}{8}ar(\triangle ABC)$

Sol:

Given: A Î”ABC$\Delta ABC$ in which AD is the medium and E is the mid-point of BD. O is the mid-point of AE.

To prove: ar(Î”BOE)=18ar(Î”ABC)$ar\left ( \Delta BOE \right )=\frac{1}{8}ar\left ( \Delta ABC \right )$

Proof: Since O is the mid-point of AE.

So, BO is the medium of Î”BAE$\Delta BAE$

Therefore, ar(Î”BOE)=12ar(Î”ABE)$ar\left ( \Delta BOE \right )=\frac{1}{2}ar\left ( \Delta ABE \right )$ . . . . (1)

Now, E is the mid-point of BD

So AE divides Î”ABD$\Delta ABD$ into two triangles of equal area.

Therefore, ar(Î”ABE)=12ar(Î”ABD)$ar\left ( \Delta ABE \right )=\frac{1}{2}ar\left ( \Delta ABD \right )$ . . . .(2)

As D is the mid-point of BC

So, ar(Î”ABD)=12ar(Î”ABC)$ar\left ( \Delta ABD \right )=\frac{1}{2}ar\left ( \Delta ABC \right )$ . . . . . (3)

Therefore, ar(Î”BOE)=12ar(Î”ABE)$ar\left ( \Delta BOE \right )=\frac{1}{2}ar\left ( \Delta ABE \right )$ [From (1)]

= 12[12ar(Î”ABCD)]$\frac{1}{2}\left [ \frac{1}{2}ar\left ( \Delta ABCD\right ) \right ]$ [From (2)]

= 14ar(Î”ABD)$\frac{1}{4}ar\left ( \Delta ABD \right )$

= 14Ã—12ar(Î”ABC)$\frac{1}{4}\times \frac{1}{2}ar\left ( \Delta ABC \right )$ [From (3)]

= 18ar(Î”ABC)$\frac{1}{8}ar\left ( \Delta ABC \right )$

Q.8: In the adjoining figure, ABCD parallelogram and O is any point on the diagonal AC.

Show that: ar.( â–³AOB$\triangle AOB$ )=ar.( â–³AOD$\triangle AOD$)

Sol:

Given: A parallelogram ABCD in which O is any point on the diagonal AC

To Prove: ar(Î”AOB)=ar(Î”AOD)$ar\left ( \Delta AOB \right )=ar\left ( \Delta AOD \right )$.

Construction: Join BD which intersects AC at P.

Proof: As diagonals of a parallelogram bisect each other therefore, OP is the median of Î”ODB$\Delta ODB$

Therefore, ar(Î”ODP)=ar(Î”OBP)$ar\left ( \Delta ODP \right )=ar\left ( \Delta OBP \right )$

Also, AP is the median of Î”ABD$\Delta ABD$

Therefore, ar(Î”ODP)=ar(Î”OBP)$ar\left ( \Delta ODP \right )=ar\left ( \Delta OBP \right )$

ar(Î”ODP)+ar(Î”ADP)=ar(Î”OBP)+ar(Î”ABP)$ar\left ( \Delta ODP \right )+ar\left ( \Delta ADP \right )=ar\left ( \Delta OBP \right )+ar\left ( \Delta ABP \right )$

Therefore, ar(Î”AOB)=ar(Î”AOD)$ar\left ( \Delta AOB \right )=ar\left ( \Delta AOD \right )$.

Q.9: P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of parallelogram ABCD. Show that PQRS is parallelogram and also show that

ar.(parallelogram PQRS)= 12$\frac{1}{2}$ * ar.(parallelogram ABCD).

Sol:

Given: ABCD is a parallelogram and P,Q,R and S are the midpoints of AB,BC,CD,DA respectively.

To prove: PQRS is a parallelogram and parallelogram PQRS= 12$\frac{1}{2}$ Parallelogram ABCD

Construction: Join AC,BD and SQ.

Proof: As S and R are the midpoints of AD and CD, so in â–³ADC$\bigtriangleup ADC$

SRâˆ¥AC$SR\parallel AC$ ( By midpoint theorem)

Also, as P and Q are the midpoints of AB and BC. So, in â–³ABC$\bigtriangleup ABC$,

PQâˆ¥AC$PQ \parallel AC$

Therefore, PQâˆ¥ACâˆ¥SR$PQ \parallel AC \parallel SR$

So, PQâˆ¥SR$PQ \parallel SR$

Similarly, we can prove SPâˆ¥RQ$SP \parallel RQ$.

Thus PQRS is a parallelogram as its opposite sides are parallel.

Since the diagonals of a parallelogram bisect each other.

So in â–³ABD$\bigtriangleup ABD$,

O is the midpoint of AC and S is the midpoint of AD.

Therefore, OSâˆ¥AB$OS\parallel AB$ (by midpoint theorem )

Similarly in â–³ABC$\bigtriangleup ABC$ we can prove that,

OQâˆ¥AB$OQ\parallel AB$ and SQâˆ¥AB$SQ\parallel AB$

Thus, ABQS is a parallelogram.

Now, â–³SPQ=12(parallelogramABQS)$\bigtriangleup SPQ =\frac{1}{2}(parallelogram \; ABQS)$ . . . . . . . (i)

Similarly, we can prove that:

â–³SPQ=12(parallelogramSQCD)$\bigtriangleup SPQ =\frac{1}{2}(parallelogram \; SQCD)$ . . . . . . . (ii)

Adding (i) and (ii) we get,

â–³SPQ+â–³SPQ=12(parallelogramABQS)+12(parallelogramSQCD)$\bigtriangleup SPQ + \bigtriangleup SPQ =\frac{1}{2}(parallelogram \; ABQS) + \frac{1}{2}(parallelogram \; SQCD)$

Therefore, (parallelogramPQRS)=12(parallelogramABCD)$(parallelogram \; PQRS)= \frac{1}{2}(parallelogram \; ABCD)$

Q.10: The given figure shows a pentagon ABCDE. EG, drawn parallel to DA, meets BA produced at G, and CF, drawn parallel to DB, meets AB produced at F. Show that ar.(pentagon ABCDE) = ar.( â–³DGF$\triangle DGF$).

Sol:

Given: ABCDE is a pentagon, EDâˆ¥DA$ED\parallel DA$ and CFâˆ¥DB$CF\parallel DB$.

To prove: ar(pentagon ABCDE) = ar(â–³DGF$\bigtriangleup DGF$)

Proof:

ar(pentagon ABCDE) = ar(â–³DBC$\bigtriangleup DBC$) +ar(â–³ADE$\bigtriangleup ADE$)+ar(â–³ABD$\bigtriangleup ABD$) . . . . . . . . (i)

Also, ar(â–³DGF$\bigtriangleup DGF$)= ar(â–³DBF$\bigtriangleup DBF$)+ar(â–³ADG$\bigtriangleup ADG$)+ â–³ABD$\bigtriangleup ABD$ . . . . . (ii)

Now, â–³DBC$\bigtriangleup DBC$ and â–³DBF$\bigtriangleup DBF$ lie on the same base and between the same parallel lines.

Therefore, ar(â–³DBC)=ar(â–³DBF)$ar(\bigtriangleup DBC)=ar(\bigtriangleup DBF)$ . . . . . . (iii)

Similarly, â–³ADE$\bigtriangleup ADE$ and â–³ADG$\bigtriangleup ADG$ lie on the same base and between the same parallel lines.

Therefore, ar(â–³ADE)=ar(â–³ADG)$ar(\bigtriangleup ADE)=ar(\bigtriangleup ADG)$ . . . . . . . (iv)

From (iii) and (iv) we have:

ar(â–³DBC)+ar(â–³ADE)=ar(â–³DBF)+ar(â–³ADG)$ar(\bigtriangleup DBC)+ ar(\bigtriangleup ADE) =ar(\bigtriangleup DBF) +ar(\bigtriangleup ADG)$

Adding ar(â–³ABD)$ar(\bigtriangleup ABD)$ on both sides

ar(â–³DBC)+ar(â–³ADE)+ar(â–³ABD)=ar(â–³DBF)+ar(â–³ADG)+ar(â–³ABD)$ar(\bigtriangleup DBC)+ ar(\bigtriangleup ADE) + ar(\bigtriangleup ABD) =ar(\bigtriangleup DBF) +ar(\bigtriangleup ADG) + ar(\bigtriangleup ABD)$

By substituting the values from (i) and (ii) we get:

ar(pentagon ABCDE) = ar(â–³DGF$\bigtriangleup DGF$)

Q.11: Prove that a median divides a triangle into two triangles of equal area.

Sol:

Let, AD is a median of â–³ABC$\bigtriangleup ABC$ and D is the midpoint of BC. AD divides â–³ABC$\bigtriangleup ABC$ in two triangles :â–³ABDandâ–³ADC$\bigtriangleup ABD \; and \; \bigtriangleup ADC$

To prove: ar(â–³ABD)=ar(â–³ADC)$ar(\bigtriangleup ABD)=ar(\bigtriangleup ADC)$

Construction: Draw ALâŠ¥BC$AL\perp BC$

Proof:

Since, D is the midpoint of BC, we have

BD=DC$BD=DC$

Multiplying with 12AL$\frac{1}{2}AL$ on both sides, we have

12Ã—BDÃ—AL=12Ã—DCÃ—AL$\frac{1}{2} \times BD\times AL=\frac{1}{2}\times DC\times AL$

Therefore, ar(â–³ABD)=ar(â–³ADC)$ar(\bigtriangleup ABD)=ar(\bigtriangleup ADC)$

Q.12: Show that a diagonal divides a parallelogram into two triangles of equal area.

Sol:

Let, ABCD be a parallelogram and BD be its diagonal.

To prove: ar(â–³ABD)=ar(â–³CDB)$ar(\bigtriangleup ABD)=ar(\bigtriangleup CDB)$

Proof:

In â–³ABDandâ–³CDB$\bigtriangleup ABD \; and \; \bigtriangleup CDB$ we have,

AB = CD (opposite sides of a parallelogram)

AD = CD (opposite sides of a parallelogram)

BD = DB (common side)

Therefore, â–³ABDâ‰…â–³CDB$\bigtriangleup ABD\cong \bigtriangleup CDB$Â (SSS criteria )

ar(â–³ABD)=ar(â–³CDB)$ar(\bigtriangleup ABD)=ar(\bigtriangleup CDB)$

Q.13: The base BC ofÂ â–³ABC$\triangle ABC$ is divided at D such that BD = 12DC$\frac{1}{2}DC$ . Prove that

ar(Â â–³ABD$\triangle ABD$) = 13Ã—ar(â–³ABC)$\frac{1}{3}\times ar(\triangle ABC)$

Sol:

Given: D is a point on BC of â–³ABC$\bigtriangleup ABC$, such that BD=12DC$BD=\frac{1}{2}DC$

To prove: ar(â–³ABD)=13ar(â–³ABC)$ar(\bigtriangleup ABD)=\frac{1}{3}ar(\bigtriangleup ABC)$

Construction: Draw ALâŠ¥BC$AL\perp BC$

Proof:

In â–³ABC$\bigtriangleup ABC$, we have :

BC = BD + DC

â‡’BD+2BD=3Ã—BD$\Rightarrow BD+2 BD =3\times BD$

Now we have:

ar(â–³ABD)=12Ã—BDÃ—AL$ar(\bigtriangleup ABD)=\frac{1}{2}\times BD\times AL$
ar(â–³ABC)=12Ã—BCÃ—AL$ar(\bigtriangleup ABC)=\frac{1}{2}\times BC\times AL$
â‡’ar(â–³ABC)=12Ã—3BDÃ—AL=3(12Ã—BDÃ—AL)$\Rightarrow ar(\bigtriangleup ABC)=\frac{1}{2}\times 3BD\times AL=3\left (\frac{1}{2}\times BD\times AL \right )$
â‡’ar(â–³ABC)=3Ã—ar(â–³ABD)$\Rightarrow ar(\bigtriangleup ABC)=3\times ar(\bigtriangleup ABD)$

Therefore, ar(â–³ABD)=13ar(â–³ABC)$ar(\bigtriangleup ABD) = \frac{1}{3} ar (\bigtriangleup ABC)$

Q.14: In the adjoining figure, the point D divides the side BC ofÂ â–³ABC$\triangle ABC$ in the ratio m : n.

Prove that:

ar(â–³ABD$\triangle ABD$):ar(â–³ADC$\triangle ADC$) = m:n.

Sol:

Given: D is a point on BC of â–³ABC$\bigtriangleup ABC$ such that BD:DC=m:n$BD:DC= m: n$

To prove: ar(â–³ABD):ar(â–³ADC)=m:n$ar(\bigtriangleup ABD):ar (\bigtriangleup ADC)= m:n$

Construction : Draw ALâŠ¥BC$AL\perp BC$

Proof:

ar(â–³ABD)=12Ã—BDÃ—AL$ar(\bigtriangleup ABD )=\frac{1}{2}\times BD\times AL$ . . . . . . (i)

ar(â–³ABD)=12Ã—DCÃ—AL$ar(\bigtriangleup ABD )=\frac{1}{2}\times DC \times AL$ . . . . . . (ii)

Dividing (i) and (ii) we have:

ar(â–³ABD)ar(â–³ABC)=12Ã—BDÃ—AL12Ã—DCÃ—AL$\frac{ar(\bigtriangleup ABD )}{ar(\bigtriangleup ABC )}=\frac{\frac{1}{2}\times BD\times AL}{\frac{1}{2}\times DC \times AL}$

=BDDC$=\frac{BD}{DC}$ = mn$\frac{m}{n}$

Therefore, ar(â–³ABD):ar(â–³ADC)=m:n$ar(\bigtriangleup ABD):ar (\bigtriangleup ADC)= m:n$

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