Advertisements
Advertisements
प्रश्न
In Fig. 10.22, the sides BA and CA have been produced such that: BA = AD and CA = AE.
Prove that segment DE || BC.
Advertisements
उत्तर
Given that, the sides BA and CA have been produced such that BA =AD and CA= AE and
given to prove DE || BC
Consider triangle BAC and , DAE
We have
BA = ADand CA = AE [∵ given in the data]
And also ∠BAC=∠DAE [ ∵vertically opposite
angles]
So, by SAS congruence criterion, we have ΔBAC ≅ ΔDAE
⇒ BC = DE and ∠DEA=∠BCA, ∠EDA ∠CBA
[Corresponding parts of congruent triangles are equal]
Now, DE and BC are two lines intersected by a transversal DB such that ∠DEA =∠BCA,
i.e., alternate angles are equal
Therefore, DE || BC
APPEARS IN
संबंधित प्रश्न
If ΔDEF ≅ ΔBCA, write the part(s) of ΔBCA that correspond to ∠F
CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ΔADE ≅ΔBCE.
Which of the following is not a criterion for congruence of triangles?
In the following figure, OA = OC and AB = BC.

Prove that: AD = CD
The following figure has shown a triangle ABC in which AB = AC. M is a point on AB and N is a point on AC such that BM = CN.
Prove that: (i) BN = CM (ii) ΔBMC ≅ ΔCNB

In the given figure, prove that: ∆ ABD ≅ ∆ ACD

In the given figure, prove that:
(i) ∆ ACB ≅ ∆ ECD
(ii) AB = ED

A is any point in the angle PQR such that the perpendiculars drawn from A on PQ and QR are equal. Prove that ∠AQP = ∠AQR.
AD and BE are altitudes of an isosceles triangle ABC with AC = BC. Prove that AE = BD.
Prove that in an isosceles triangle the altitude from the vertex will bisect the base.
