In - ABC - AB - AC and BD - AC prove that BD 2- CD 2-2 AC-CD -

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Pythagoras Theorem

In Δ ABC , AB...

Question

In ΔABC, AB = AC and BD ⊥ AC prove that BD² + CD² = 2AC.CD.

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Δ A B C, A B = A C

B D ⊥ A C

B D^{2} - C D^{2} = 2 C D \times A D

A B C

A B = A C

B D

A C

{B D}^{2} - {C D}^{2} = 2 C D \times A D

B D^{2} + C D^{2} = 2 A C \times C D

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