Home
Class 10
MATHS
In Figure altitudes AD and CE of triangl...

In Figure altitudes `AD` and `CE` of `triangle ABC` intersect each other at the point P. Show that: (i) `DeltaA E P~ DeltaC D P` (ii) `DeltaA B D ~ DeltaC B E` (iii) `DeltaA E P~ DeltaA D B` (iv) `DeltaP D C~ DeltaB E C`

Text Solution

Verified by Experts

(i) in figure, `angleAEP = angleCDP " " (each = 90^(@)`
and ` angleAPE = angleCPD` ( vertically opposite angles)
`Rightarrow " " triangleAEP ~ triangleCDP` (by AAA similarity critertion)
Hence proved
`(ii) figure , angleADB= angleCEB " " (each = 90^(@))`
`and " " angleABD= angleCBE` ( by AAA similarity criterion) Hence proved.
( iii) in figure, ` angleAEP = angleADB " " (each 90^(@))`
` and anglePAE = angleBAD` (common angle)
`Rightarrow " " triangleAEP ~ triangleADB` ( by AAA similarity criterion ) Hence proved.
(iv) in figure , `anglePDC = angleBEC ( each = 90^(@))`
and ` anglePCD = angleBCE` ( common angle)
` Rightarrow " " trianglePDC ~ triangleBEC` ( by AAA similarity criterion) Hence proved.
Promotional Banner

Topper's Solved these Questions

  • TRIANGLES

    NAGEEN PRAKASHAN|Exercise Exercise 6a|23 Videos
  • TRIANGLES

    NAGEEN PRAKASHAN|Exercise Eercise 6b|1 Videos
  • TRIANGLES

    NAGEEN PRAKASHAN|Exercise Revision Exercise Long Questions|4 Videos
  • STATISTICS

    NAGEEN PRAKASHAN|Exercise Revision Exercise Long Answer Questions|5 Videos
  • VOLUME AND SURFACE AREA OF SOLIDS

    NAGEEN PRAKASHAN|Exercise Revisions Exercise Long Answer Questions|5 Videos

Similar Questions

Explore conceptually related problems

In Figure altitudes AD and CE of DABC intersect each other at the point P. Show that:(i) DeltaA E P~ DeltaC D P (ii) DeltaA B D ~DeltaC B E (iii) DeltaA E P~ DeltaA D B (iv) DeltaP D C ~DeltaB E C

In Figure, two chords AB and CD intersect each other at the point P. Prove that: (i) DeltaA P C~ DeltaD P B (ii) A P*P B=C P*D P

In Figure two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that (i) DeltaP A C DeltaP D B (ii) P AdotP B=P CdotP D

In figure, if DeltaA B E~=DeltaA C D , show that DeltaA D E~ DeltaA B C .

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that DeltaA B E DeltaC F B .

Two sides AB and BC and median AM of one triangle ABC are respectively equal to side PQ and QR and median PN of DeltaA B C~=DeltaP Q R (see Fig. 7.40). Show that: (i) DeltaA B M~=DeltaP Q N (ii) DeltaA B C~=DeltaP Q R

In figure ABC and AMP are two right triangles, right angles at B and M respectively. Prove that(i) DeltaA B C~ DeltaA M P (ii) (C A)/(P A)=(B C)/(M P)

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that a r" "(A P B)" "xx" "a r" "(C P D)" "=" "a r" "(A P D)" "xx" "a r" "(B P C)dot

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that D P = B Q . Show that: (i) DeltaA P D~= DeltaC Q B (ii) A P = C Q (iii) DeltaA B C (iv) A Q = C P (v) APCQ is a parallelogram.

DeltaA B C and DeltaD B C are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that (i) \ DeltaA B D~=DeltaA C D (ii) DeltaA B P~=DeltaACP (iii) AP bisects ∠ A as well as ∠ D (iv) AP is the perpendicular bisector of BC