$Δ D A P ≅ Δ E B P$

True

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False

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Solution

The correct option is A True
In triangles $D A P$ and $E B P,$
$A P = P B$ (Since, P is mid point of AB)
$∠ B A D = ∠ A B E$ ......... (Given)
$∠ E P A = ∠ D P B$ ........ (Given)
$∠ A P D = ∠ B P E$ ....... [Since $∠ E P D$ is a part of both the triangles]

Hence According to ASA Congruency Criterion
$△ D A P ≅ △ E B P$

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Similar questions

A B

is a line segment and

P

is its mid-point.

D

and

E

are points on the same side of

A B

such that

∠ B A D = ∠ A B E

and

∠ E P A = ∠ D P B

. Show that

Δ D A P ≅ Δ E B P

$A B$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of
$A B$ such that $∠ B A D = ∠ A B E and ∠ E P A = ∠ D P B$ (See the given figure). Show that
$Δ D A P ≅ Δ E B P$

Q. Two angles can have exactly five points in common.

True
False

X = -5 can this be expressed as a linear equation in two variables?

True

False

State true\false:

3 x + 2 y = 5, 2 x - 3 y = 7

is an example of consistent pair of equations.

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ΔDAP≅ΔEBP

The correct option is A TrueIn triangles DAP and EBP,AP=PB (Since, P is mid point of AB) ∠BAD=∠ABE ......... (Given)∠EPA=∠DPB ........ (Given)∠APD=∠BPE ....... [Since ∠EPD is a part of both the triangles]Hence According to ASA Congruency Criterion △DAP≅△EBP

$Δ D A P ≅ Δ E B P$