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# D is a point on the side BC of a triangle ABC such that ∠ADC = ∠ BAC. Show that CA^{2} = CB.CD

**Solution:**

We know that if two triangles are similar, then their corresponding sides are proportional.

In ΔABC and ΔDAC

∠BAC = ∠ADC (Given in the statement)

∠ACB = ∠ACD (Common angles)

⇒ ΔABC ∼ ΔDAC (AA criterion)

If two triangles are similar, then their corresponding sides are proportional

⇒ CA / CD = CB / CA

⇒ CA^{2} = CB × CD

Hence, proved.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 6

**Video Solution:**

## D is a point on the side BC of a triangle ABC such that ∠ADC = ∠ BAC. Show that CA² = CB.CD

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 Question 13

**Summary:**

D is a point on the side BC of a triangle ABC such that ∠ADC = ∠ BAC. We have proved that CA^{2} = CB.CD.

**☛ Related Questions:**

- E is a point on the side AD produced of a parallelogram ABCD and BE intersect CD at F. Show that ΔABE ~ ΔCFB.
- In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ΔABC ~ ΔAMP (ii) CA/PA = BC/MP.
- CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that: (i) CD/GH =AC/FG (ii) ∆DCB ~ ∆HGE (iii) ∆DCA ~ ∆HGF
- In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥AC, prove that ∆ ABD ~ ∆ ECF.

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