Question

baby is a line segment and P is its midpoint. e and e are points on the same sides of a b such that angle B = triangle ABC and angle wpa equal to angle the PV show that triangle TP the AP congruent evp and equals to be​

Answer 1

Step-by-step explanation:

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$\huge \red {\boxed {\boxed {\mathbb {QUESTION}}}}$

AB is aline segment And P is its mid point. D and E are point on the same side of AB such that < BAD = < ABE and <EPA = < DPB. Show that. 1.∆ DAP is congruent to ∆EPB. 2.AD =BE.

$\huge \red {\boxed {\boxed {\mathbb {ANSWER}}}}$

To Prove : 1.∆ DAP is congruent to ∆EPB. 2.AD =BE.

∠ EPA = ∠ DPB (Given)

⇒ ∠ EPA + ∠DPE = ∠ DPB + ∠DPE

Therefore∠APD = ∠BPE

Consider Triangles DAP and EBP

AP = BP (Given P is midpoint of AB)

∠BAD = ∠ABE (Given)

∠APD = ∠BPE (Proved)

Hence Triangle DAP is congruent to Triangle EBP (By ASA congruence rule)

⇒ AD = BE (CPCT)

Hence Proved.

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Answer 2

Answer:

$Step-by-step explanation:</p><p></p><p>\huge {\underline{ \bf \red{HEYA\:MATE}}}HEYAMATE</p><p></p><p>\huge \red {\boxed {\boxed {\mathbb {QUESTION}}}}QUESTION</p><p></p><p>AB is aline segment And P is its mid point. D and E are point on the same side of AB such that < BAD = < ABE and <EPA = < DPB. Show that. 1.∆ DAP is congruent to ∆EPB. 2.AD =BE.</p><p></p><p>\huge \red {\boxed {\boxed {\mathbb {ANSWER}}}}ANSWER</p><p></p><p>To Prove : 1.∆ DAP is congruent to ∆EPB. 2.AD =BE.</p><p></p><p>∠ EPA = ∠ DPB (Given)</p><p></p><p>⇒ ∠ EPA + ∠DPE = ∠ DPB + ∠DPE</p><p></p><p>Therefore∠APD = ∠BPE</p><p></p><p>Consider Triangles DAP and EBP</p><p></p><p>AP = BP (Given P is midpoint of AB)</p><p></p><p>∠BAD = ∠ABE (Given)</p><p></p><p>∠APD = ∠BPE (Proved)</p><p></p><p>Hence Triangle DAP is congruent to Triangle EBP (By ASA congruence rule)</p><p></p><p>⇒ AD = BE (CPCT)</p><p></p><p>Hence Proved.</p><p></p><p>\huge{\red{\tt Hope \: it \: helps \: you}}Hopeithelpsyou</p><p></p><p>\color{lightgreen}{{MARK IT AS BRAINLIEST!! }}MARKITASBRAINLIEST!!</p><p></p><p>$