In the given figure- - CAB - 90o and AD - BC- Show that - BDA - BAC- If AC - 75 cm- AB - 1 m and BC - 1-25 m- find AD-

In CAB and ADB ∠ CAB = ∠ ADB = 90^o ∠ CBA = ∠ ABD (Common) By, AA similarity, CAB ∼ ADB So, CA/AD = CB/AB 75/AD = 125/100 AD = 75 × 100/12 ...

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Intersection between Secants

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In the given figure, $∠ C A B = 90^{o}$ and $A D ⊥ B C$ . Show that $Δ B D A \sim Δ B A C$ . If AC = 75 cm, AB = 1 m and BC = 1.25 m, find AD.

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