Question

22. Dis a point on the side BC of a triangle ABC such that ZADC - 2 BAC.
Show that ca? - CB.CD.
A
B
D​

Answer 1

Answer:

In ΔADC and ΔBAC,

∠ADC = ∠BAC (Given)

∠ACD = ∠BCA (Common angle)

∴ ΔADC ~ ΔBAC (By AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.

∴ CA/CB =CD/CA

⇒ CA2 = CB.CD.

An alternate method:

Given in ΔABC, ∠ADC = ∠BAC

Consider ΔBAC and ΔADC

∠ADC = ∠BAC (Given)

∠C = ∠C (Common angle)

∴ ΔBAC ~ ΔADC (AA similarity criterion)

⇒ \frac{AB}{AD} = \frac{CB}{CA} = \frac{CA}{CD}

Consider, \frac{CB}{CA} = \frac{CA}{CD}

∴ CB x CD = CA²