ABC is a triangle, D is a point on AB such that $A D = \frac{1}{4} A B$ and E is a point on AC such that $A E = \frac{1}{4}$ AC. Prove that DE = $\frac{1}{4} B C$

Q. State true or false:

△ A B C

D, E

and

F

are mid-points of sides

A B, B C

and

A C

respectively. Hence

A E

and

D F

bisect each other

Q. In a triangle

A B C, A B = A C

and

D

is a point on side

A C

such that

B C^{2} = A C \times C D

prove that

B D = B C

Q. In a triangle ABC, AB = AC and D is a point on side AC such that BC = AC

\times

CD. Prove that BD = BC.

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State true or false:In triangle ABC, D is a point in side AB such that AB=4AD and E is a point in AC such that AC=4AE. Hence, BC:DE=4:1.

The correct option is A TrueGiven ABAD=4and ACAE=4Hence, ABAD=ACAEIn △ADE and △ABC,∠DAE=∠BAC (Common)ABAD=ACAE (Given)Thus, △ADE∼△ABC (SAS rule)Hence, BCDE=ABAD=41 (Corresponding Side)

State true or false:
In triangle $A B C$ , $D$ is a point in side $A B$ such that $A B = 4 A D$ and $E$ is a point in $A C$ such that $A C = 4 A E$ . Hence, $B C : D E = 4 : 1$ .