In Fig. 6.38, altitudes AD and CE of Δ ABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC
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 to as the AA similarity criterion for two triangles.
In ΔAEP and ΔCDP
∠AEP = ∠CDP = 90º
[∵ CE ⊥ AB and AD ⊥ BC; altitudes]
∠APE = ∠CPD (Vertically opposite angles)
⇒ ΔAEP ~ ΔCPD (AA criterion)
(ii) 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 ΔABD and ΔCBE
∠ADB = ∠CEB = 90º
∠ABD = ∠CBE (Common angle)
⇒ ΔABD ~ ΔCBE (AA criterion)
(iii) 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 ΔAEP and ΔADB
∠AEP = ∠ADB = 90º
∠PAE = ∠BAD (Common angle)
⇒ ΔAEP ~ ΔADB (AA criterion)
(iv) If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
This is referred to as the AA similarity criterion for two triangles.
In ΔPDC and ΔBEC
∠PDC = ∠BEC = 90º
∠PCD = ∠BCE (Common angle)
⇒ ΔPDC ~ ΔBEC (AA criterion)
☛ Check: NCERT Solutions for Class 10 Maths Chapter 6
Video Solution:
In Fig. 6.38, altitudes AD and CE of Δ ABC intersect each other at the point P. Show that: (i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE (iii) ΔAEP ~ ΔADB (iv) ΔPDC ~ ΔBEC
NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.3 Question 7
Summary:
If in the above figure altitudes, AD and CE of Δ ABC intersect each other at the point P, it is proved that (i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE (iii) ΔAEP ~ ΔADB and (iv) ΔPDC ~ ΔBEC
☛ 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.