Home
Class 10
MATHS
In figure, if DeltaA B E~=DeltaA C D, sh...

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

Text Solution

Verified by Experts

`(##X_06_E03_06_s10##)`
Given, `△ABE≅△ACD`
Hence, `AB=AC` by CPCT. .....(1)
[CPTC: corresponding parts of congruent triangles] ...
Promotional Banner

Topper's Solved these Questions

  • TRIANGLES

    NCERT ENGLISH|Exercise Exercise 6.2|10 Videos

  • TRIANGLES

    NCERT ENGLISH|Exercise SOLVED EXAMPLES|14 Videos

  • TRIANGLES

    NCERT ENGLISH|Exercise Exercise 6.5|17 Videos

  • SURFACE AREAS AND VOLUMES

    NCERT ENGLISH|Exercise Exercise 13.3|9 Videos

Similar Questions

Explore conceptually related problems

Delta\ A B C\ a n d\ Delta\ A B D are on a common base A B ,\ a n d\ A C\ =\ B D\ a n d\ B C\ =\ A D as shown in Fig. 18. Which of the following statements is true? DeltaA B C~=DeltaA B D DeltaA B C~=DeltaA D B DeltaA B C~=DeltaB A D

In figure Cm and RN are respectively the medians of DeltaA B C and DeltaP Q R . If DeltaA B C ~DeltaP Q R , prove that: (i) DeltaA M C ~DeltaP N R (ii) (C M)/(R N)=(A B)/(P Q) (ii) DeltaC M B ~DeltaR N Q

In Figure, if /_\ A B E~= /_\A C D , prove that /_\ A D E ~ /_\ A B C .

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

In Figure altitudes AD and CE of A B C 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

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of DeltaP Q R . Show that DeltaA B C~ DeltaP Q R .

In figure.20, DeltaA B C is isosceles with A B=A Cdot State if DeltaA B C~=DeltaA C Bdot If yes, state three relations that you use to arrive at your answer.

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 .

In figure, A B|| D C\ a n d\ A B=D Cdot (i)Is DeltaA C D~=DeltaC A B ? (ii)State the three pairs of matching parts used to answer (iii) Which angle is equal to /_C A D\ ? (iv) Does it follow from (iii) that A D || B C ?

ABC is a right triangle right angled at A. BCED, ACFG and ABMN aresquares on the sides BC, CA and AB respectively. Line segment A X_|_D E meets BCat Y. Show that:(i) DeltaM B C~=DeltaA B D (ii) a r\ (B Y X D)\ =\ 2\