Home
Class 10
MATHS
In Fig. Altitudes AD and CE of Delta AB...

In Fig. Altitudes AD and CE of `Delta ABC ` intersect each other at the point P . Show that :
(i) `DeltaAEP~DeltaCDP`
(ii) `DeltaABD~DeltaCBE`
(iii) `DeltaAEP ~DeltaADB`
(iv) `Delta PDC ~DeltaBEC`

Text Solution

Verified by Experts

The correct Answer is:
(i) `Delta AEP ~DeltaCDP` (`:'` A - A similarity criterian) , (ii) `:. Delta ABD ~ Delta CBE` ,
(iii) `Delta AEP~ DeltaADB` (By A A A similarity criteria) . , (iv) `:. Delta PDC ~ Delta BEC` (By A A A cruterion).
Promotional Banner

Topper's Solved these Questions

  • TRIANGLES

    CPC CAMBRIDGE PUBLICATION|Exercise Exercise 2.4|9 Videos

  • TRIANGLES

    CPC CAMBRIDGE PUBLICATION|Exercise Exercise 2.5|17 Videos

  • TRIANGLES

    CPC CAMBRIDGE PUBLICATION|Exercise Exercise 2.2|10 Videos

  • SURFACE AREAS AND VOLUMES

    CPC CAMBRIDGE PUBLICATION|Exercise EXERCISE 15.4|5 Videos

Similar Questions

Explore conceptually related problems

In Fig , altitudes AD and CE of triangle ABC intersect each other at the point P. show that : DeltaAEP ~DeltaADB

In Fig , altitudes AD and CE of triangle ABC intersect each other at the point P. show that : DeltaABD ~DeltaCBE

In Fig , altitudes AD and CE of triangle ABC intersect each other at the point P. show that : DeltaPDC ~DeltaBEC

In Fig , altitudes AD and CE of triangle ABC intersect each other at the point P. show that : triangleAEP ~triangleCDP

In Fig. two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that (i) Delta PAC ~ Delta PDB , (ii) PA . PB = PC . PD

In the given Figure AB and CD are intersecting at ‘O’, OA = OB and OD = OC. Show that (i) Delta AOD ~= DeltaBOC and (ii) AD || BC .

In an isosceles triangle ABC, with AB = AC, the bisectors of /_ B and /_ C intersect each other at O. Join A to O. Show that : (i) OB = OC (ii) AO bisects /_A

ABC is a triangle in which altitudes BD and CE to sides AC and AB are equal (see figure) . Show that (i) DeltaABD ~= DeltaACE (ii) AB = AC i.e., ABC is an isosceles triangle.

In right triangle ABC, right angle is at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that : (i) DeltaAMC ~= DeltaBMD (ii) /_DBC is a right angle (iii) Delta DBC ~= DeltaACB (iv) CM = 1/2 AB .