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Converse of Basic Proportionality Theorem

Question 7 ii...

Question

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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Solution

(iii) $I n Δ A E P a n d Δ A D B$ ,
$∠ A E P = ∠ A D B (E a c h 90^{\circ})$
$∠ P A E = ∠ D A B (C o m m o n)$
Hence, by using AA similarity criterion,
$Δ A E P \sim Δ A D B$

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Similar questions

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

Q. Question 7 (iv)
In the below figure, altitudes AD and CE of

Δ A B C

intersect each other at the point P. Show that

Δ P D C \sim Δ B E C

Q. In the figure, altitudes

A D

C E

△ A B C

intersect each other at the point

P

△ A E P \sim △ C D P

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Converse of Basic Proportionality Theorem

Standard X Mathematics

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