Question

AB is a line segment and P is its mid-point. D and E are points on the same side of AB ,Such that < = < and < = < . Show that PD = PE. ​

Answer 1

Answer:

$\huge\bf\underline{\underline{\pink{A}\orange{N}\red{S}\green{W}\purple{E}\blue{R}}}$

To Prove :

∆ DAP is congruent to ∆EPB. 2.AD =BE.

∠ EPA = ∠ DPB (Given)

⇒ ∠ EPA + ∠DPE = ∠ DPB + ∠DPE

∴∠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)

⇒ PD = PE (CPCT)

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Answer 2

Answer:

1. by congruent property these are equal

Step-by-step explanation:

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