# 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

**Solution:**

(i) If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

This is referred as AA similarity criterion for two triangles.

In ΔABC and ΔAMP

∠ABC = ∠AMP = 90º

∠BAC = ∠MAP (Common angle)

Thus, ΔABC ∼ ΔAMP (AA criteria)

(ii) As we know that the ratio of any two corresponding sides in two similar triangles is always the same,

In ΔABC and ΔAMP

CA/PA = BC/MP [∵ ΔABC ∼ ΔAMP]

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

**Video Solution:**

## 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

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.3 Question 9

**Summary:**

In the above figure, ABC and AMP are two right triangles, right-angled at B and M respectively. Hence it is proved that ΔABC ~ ΔAMP and CA/PA = BC/MP.

**☛ Related Questions:**

- 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.
- Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆ PQR (see Fig. 6.41). Show that ∆ ABC ~ ∆ PQR.
- D is a point on the side BC of a triangle ABC such that ∠ADC = ∠ BAC. Show that CA^2 = CB.CD.

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