Skip to content
Philoid
Browse Saved
Back to chapter
Maths
6. Triangles
Home · Class 10 · Maths · NCERT - Mathematics · 6. Triangles
Prev
Next
Q8 of 65 Page 138

E is a point on the side AD produced of aparallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ Δ CFB

In ΔABE and ΔCFB,


∠A = ∠C (Opposite angles of a parallelogram)


∠AEB = ∠CBF (Alternate interior angles because AE || BC)


Therefore,


ΔABE ΔCFB (By AA similarity)


More from this chapter

All 65 →
6

In Fig. 6.37, if Δ ABE ≅Δ ACD, show that
Δ ADE ~ Δ ABC.

7

In Fig. 6.38, altitudes AD and CE of Δ ABCintersect each other at the point P. Showthat:

(i) Δ AEP ~ Δ CDP


(ii) Δ ABD ~ Δ CBE


(iii) Δ AEP ~ Δ ADB


(iv) Δ PDC ~ Δ BEC


9

In Fig. 6.39, ABC and AMP are two righttriangles, right angled at B and Mrespectively. Prove that:

(i) Δ ABC ~ Δ AMP


(ii)


10

CD and GH are respectively the bisectorsof ∠ ACB and ∠ EGF in such a way that D and H lieon sides AB and FE of Δ ABC and Δ EFGrespectively. If Δ ABC ~ Δ FEG, show that:

(i)


(ii) Δ DCB ~ Δ HGE


(iii) Δ DCA ~ Δ HGF

Questions · 65
6. Triangles
1 2 3 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 2 3 4 5 6 7 8 9 10
Back to chapter
ADVERTISEMENT
About Contact Privacy Terms
Philoid · 2026
  • Home
  • Search
  • Browse
  • Quiz
  • Saved