In the figure, altitudes $A D$ and $C E$ of $△ A B C$ intersect each other at the point $P$ .Show that $△ A E P \sim △ C D P$

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Solution

Given: $△ A B C$ and altitude $A D$ and $C E$ of triangle intersects each other at the point $P$ .

To prove: $△ A E P \sim △ C D P$

Proof:In $△ A E P$ and $△ C D P$

$∠ A E P = ∠ C D P$ since $A D$ and $C E$ are altitudes.

$∠ A P E = ∠ C P D$ (vertically opposite angles)

Hence, $△ A E P \sim △ C D P$ by AAA similarity

Hence proved.

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Similar questions

Q. In Fig, altitudes

A D

and

C E

△ A B C

intersect each other at the point

P .

Show that:
(i)

△ A E P \sim △ C D P

(ii)

△ A B D \sim △ C B E

(iii)

△ A E P \sim △ A D B

(iv)

△ P D C \sim △ B E C

Question 7 (i)
In the figure, altitudes AD and CE of $Δ A B C$ intersect each other at the point P. Show that

(i) $Δ A E P \sim Δ C D P$

Question 7 (i)
In the figure, altitudes AD and CE of $Δ A B C$ intersect each other at the point P. Show that

(i) $Δ A E P \sim Δ C D P$

In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:

(i) ΔAEP ∼ ΔCDP

(ii) ΔABD ∼ ΔCBE

(iii) ΔAEP ∼ ΔADB

(v) ΔPDC ∼ ΔBEC

Question 7 (iii)
In the below figure, altitudes AD and CE of $Δ A B C$ intersect each other at the point P. Show that

(iii) $Δ A E P \sim Δ A D B$

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In the figure, altitudes AD and CE of △ABC intersect each other at the point P.Show that △AEP∼△CDP

Given:△ABC and altitude AD and CE of triangle intersects each other at the point P.To prove:△AEP∼△CDPProof:In △AEP and △CDP∠AEP=∠CDP since AD and CE are altitudes.∠APE=∠CPD (vertically opposite angles)Hence, △AEP∼△CDP by AAA similarityHence proved.

In the figure, altitudes $A D$ and $C E$ of $△ A B C$ intersect each other at the point $P$ .Show that $△ A E P \sim △ C D P$