CLASS 10 CBSE

1. CH 6
TRIANGLES
EX 6.3 CONTINUE

2. 7. In the figure, altitudes AD and CE of A ABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC
HOME WORK
Sol: We have a ΔABC in which altitude AD and CE intersect each other at P.
⇒∠D = ∠E = 90° ...(1)
(i) In ΔEAP and ΔCDP
∠AEP = ∠CDP [From (1)]
∠EPA = ∠DPC [Vertically opp. angles]
∴Using AA similarity, we get
ΔEAP and ΔCDP
(ii) In ΔABD and ΔCBE
∠ADB = ∠CEB [From (1)]
Also ∠ABD = ∠CBE [Common]
∴Using AA similarity, we have
ΔABD and ΔCBE

3. (iii) In ΔAEP and ΔADB
∠AEP = ∠ADB [From (1)]
Also ∠EAP = ∠DAB [Common]
∴Using AA similarity, we have
ΔAEP and ΔADB
(iv) In ΔPDC and ΔBEC
∠PDC = ∠BEC [From (1)]
And ∠DCP = ∠ECB [Common]
∴Using AA similarity, we get
ΔPDC and ΔBEC

4. 8. E is a point on the side AD produced of a parallelogram. ABCD and BE intersects CD at F.
Show that ΔABE ~ ΔCFB.
Sol.
We, have a parallelogram ABCD in which AD is produced to E and
BE is joined such that BE intersects CD at F.
Now, in ΔABE and ΔCFB
∠BAE = ∠FCB
[Opp. angles of a || gm are always equal]
∠AEB = ∠CBF
[∵Parallel sides are intersected by the transversal BE]
Now, using AA similarity, we have
ΔABE and ΔCFB

5. 9. In the figure, ABC and AMP are two right triangles, right angle at B and M respectively. Prove that:
Sol. We have ΔABC, right angled at B and ΔAMP, right
angled at M.
∴∠B = ∠M = 90° ...(1)
(i) In ΔABC and ΔAMP
∠ABC = ∠AMP [From (1)]
And ∠BAC = ∠MAP [Common]
(ii) ΔABC ~ ΔAMP [As proved above]
∴ Their corresponding sides are proportional.
𝐶𝐴
𝑃𝐴
=
𝐵𝐶
𝑀𝑃
(i) ΔABC ~ ΔAMP (ii)
𝐶𝐴
𝑃𝐴
=
𝐵𝐶
𝑀𝑃
∴ Using AA similarity, we have
ΔABC ~ ΔAMP

6. 10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on
sides AB and FE of ΔABG and ΔEFG respectively. If ΔABC ~ ΔFEG, show that:
Sol. We have two similar ΔABC and ΔFEG such that CD and GH are the
bisectors of ∠ACB and ∠FGE respectively.
(i) In ΔACD and ΔFGH
∠CAD = ∠GFH ...(1)[ΔABC ~ ΔFEG ∴∠A =∠F]
since ΔABC ~ ΔFEG [Given]
∴∠C = ∠G ⇒ ∴
1
2
∠ACB =
1
2
∠FGE
∴∠ACD ~ ∠FGH ...(2)
From (1) and (2),
ΔACD ~ ΔFGH
∴ Their corresponding sides are proportional, ∴
𝐶𝐷
𝐺𝐻
=
𝐴𝐶
𝐹𝐺
(ii) In ΔDCB and ΔHGE
∠DBC = ∠HEG ...(1)[ΔABC ~ ΔFEG ∴∠B = E]
Again, ΔABC ~ ΔFE
∴ ∠ACB = ∠FGE ⇒ ∴
1
2
∠ACB =
1
2
∠FGE
∴∠DBC = ∠HEG...(2)
From (1) and (2), we get
ΔDCB ~ ΔHGE

7. (iii) In ΔDCA and ΔHGF
∠DAC = ∠HGE ...(1) [ΔABC ~ ΔFEG ∴∠CAB = ∠GFE ⇒ ∠CAD = ∠GFF ∴∠DAC = ∠HFG]
Also ΔABC ~ ΔFEG ∴ΔACB ~ ΔFGE
∴ ΔDCA ~ ΔHGF

8. 11. In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC.
If AD ⊥ BC an EF ⊥ AC, prove that ΔABD ~ ΔECF,
Sol. We have an isosceles ΔABC in which AB = AC.
In ΔABD and ΔECF
AB = AC [Given]
⇒Angles opposite to them are equal
∴∠ACB = ∠ABC
⇒∠ECF = ∠ABD ..(1)
Again AD ⊥ BC and EF ⊥ AC
⇒∠ADB = ∠EFC = 90° ...(2)
From (1) and (2), we have
ΔABD ~ ΔECF [AA criteria of similarity]