Question
$ABC$ and $ADC$ are two right triangles with common hypotenuse $AC$. Prove that $\angle C A D = \angle C B D.$
✓
Answer
Given: ABC and ADC are two right triangles with common hypotenuse $AC.$
To prove: $\angle C A D = \angle C B D$
Proof : AC is the common hypotenuse and $ABC$ and $ADC$ are two right triangles.
$\therefore \angle A B C = 90 ^ { \circ } = \angle A D C$
$\Rightarrow$ Both the triangles are in the same semi-circle
$\therefore$ Points $A, B, D$ and $C$ are concyclic
$\therefore DC$ is a chord
$\therefore \angle C A D = \angle C B D$ |$\because$ Angles in the same segment are equal.
To prove: $\angle C A D = \angle C B D$
Proof : AC is the common hypotenuse and $ABC$ and $ADC$ are two right triangles.
$\therefore \angle A B C = 90 ^ { \circ } = \angle A D C$
$\Rightarrow$ Both the triangles are in the same semi-circle
$\therefore$ Points $A, B, D$ and $C$ are concyclic
$\therefore DC$ is a chord
$\therefore \angle C A D = \angle C B D$ |$\because$ Angles in the same segment are equal.
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