two right triangles ABC and DBC …
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Sia ? 6 years, 4 months ago
Given: Right triangles {tex}\triangle ABC{/tex} and {tex}\triangle DBC{/tex} are drawn on the same hypotenuse BC on the same side of BC.
Also, AC and BD intersect at P.
To Prove: {tex}A P \times P C = B P \times P D{/tex}
Proof: In {tex}\triangle BAP{/tex} and {tex}\triangle CDP{/tex}, we have
{tex}\angle B A P = \angle C D P = 90 ^ { \circ }{/tex}
{tex}\angle B P A = \angle C P D{/tex} (vertically opposite angles)
{tex}\therefore \quad \triangle B A P \sim \triangle C D P{/tex} [by AA-similarity]
{tex}\therefore \quad \frac { A P } { D P } = \frac { B P } { C P }{/tex}
{tex}\Rightarrow A P \times C P = B P \times D P{/tex}
{tex}\Rightarrow A P \times P C = B P \times P D{/tex}
Hence, {tex}A P \times P C = B P \times P D{/tex}.
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