You visited us 1 times! Enjoying our articles? Unlock Full Access!

Converse of Basic Proportionality Theorem

In Δ ABC, AL ...

Question

In $Δ A B C$ , AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that:
(i) $Δ O M A \sim Δ O L C$
(ii) $\frac{O A}{O C} = \frac{O M}{O L}$

Open in App

Solution

(i) in ΔOMA and ΔOLC,

∠AOM = ∠COL (Vertically opposite angles)

∠OMA = ∠OLC (90 each)

ΔOMA∼ΔOLC (A-A similarity)

(ii) Since, ΔOMA∼ΔOLC by A-A similarity, then

$fraction numerator O M over denominator O L end fraction equals fraction numerator O A over denominator O C end fraction equals fraction numerator M A over denominator L C end fraction$ (corresponding sides of similar triangles are proportional)
$fraction numerator O A over denominator O C end fraction equals fraction numerator O M over denominator O L end fraction$

Hence proved

Suggest Corrections

Similar questions

Q. In ∆ABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that:

(i) ∆OMA ∼ ∆OLC
(ii)

\frac{OA}{OC} = \frac{OM}{OL}

Q. In

△ A B C, A L

, and

C M

are the perpendiculars from the vertices

A

and

C

B C

and

A B

respectivaly. If

A L

and

C M

intersect at

O,

Prove that:

(i) △ O M A \sim △ O L C

(i i) \frac{O A}{O C} = \frac{O M}{O L}

Q. A point O is taken inside an equilateral

Δ A B C

. If

O M ⊥ A C

O L ⊥ B C

and

O N ⊥ A B

such that OL = 14 cm, OM = 10 cm and ON = 6 cm, find the area of

Δ A B C

D A, C B, O M

are each perpendicular to line segment

A B

where

O

is the point of intersection of

A C

and

D B

.If

O A = 2.4

cm,

O C = 3.6

cm then find the ratio of

A M : B M

Triangle ABC is constructed with AB = 4 cm, ∠CAB = $45^{0}$ and BC = 2.92 cm. Two perpendicular bisectors are drawn of side AB and BC intersecting at point O. The length of OA is:

Join BYJU'S Learning Program

In ΔABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that: (i) ΔOMA∼ΔOLC (ii) OAOC=OMOL

(i) in ΔOMA and ΔOLC, ∠AOM = ∠COL (Vertically opposite angles) ∠OMA = ∠OLC (90 each) ΔOMA∼ΔOLC (A-A similarity) (ii) Since, ΔOMA∼ΔOLC by A-A similarity, then (corresponding sides of similar triangles are proportional) Hence proved

In $Δ A B C$ , AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that:
(i) $Δ O M A \sim Δ O L C$
(ii) $\frac{O A}{O C} = \frac{O M}{O L}$