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

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Solution

$I t i s g i v e n t h a t, ∠ E P A = ∠ D P B ∠ E P A + ∠ E P D = ∠ D P B + ∠ E P D [a d d i n g ∠ E P D o n b o t h s i d e s] ∠ A P D = ∠ B P E \dots (i) I n Δ D A P a n d Δ E B P, A P = B P (P i s a m i d p o i n t o f A B) ∠ A P D = ∠ B P E (F r o m (i)) ∠ B A D = ∠ A B E (G i v e n) ∴ Δ D A P ≅ Δ E B P (B y A S A c o n g r u e n c e r u l e)$

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Similar questions

Q. AB is a line segment and P is its mid-point. D and E are points on the same side of
AB such that

∠ B A D = ∠ A B E a n d ∠ E P A = ∠ D P B

(See the given figure). Show that

Δ D A P ≅ Δ E B P

Question 7 (i)
AB is a line segment and P is its mid-point. D and E are points on the same side of
AB such that $∠ B A D = ∠ A B E a n d ∠ E P A = ∠ D P B$ (See the given figure). 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
$Δ D A P ≅ Δ E B P$

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

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that

(i) ΔDAP ≅ ΔEBP

(ii) AD = BE

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AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD=∠ABEand∠EPA=∠DPB (See the given figure). Show that ΔDAP≅ΔEBP

It is given that,∠EPA=∠DPB∠EPA+∠EPD=∠DPB+∠EPD [adding ∠EPD on both sides]∠APD=∠BPE…(i)In ΔDAP and ΔEBP,AP=BP (P is a midpoint of AB)∠APD=∠BPE (From (i))∠BAD=∠ABE (Given)∴ΔDAP≅ΔEBP (By ASA congruence rule)

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