Advertisements
Advertisements
प्रश्न
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that
- ΔDAP ≅ ΔEBP
- AD = BE
Advertisements
उत्तर
We have, P is the mid-point of AB.
∴ AP =BP
∠EPA = ∠DPB ...[Given]
Adding ∠EPD to both sides, we get:
∠EPA + ∠EPD = ∠DPB + ∠EPD
⇒ ∠APD = ∠BPE
i. Now, in △DAP and △EBP, we have
∠PAD = ∠PBE ...[∵ ∠BAD = ∠ABE]
AP = BP ...[Proved above]
∠DPA = ∠EPB ...[Proved above]
∴ △DAP ≌ △EBP ...[By ASA congruency]
ii. Since △DAP ≌ △EBP
⇒ AD = BE ...[By Corresponding Parts of Congruent Triangles]
APPEARS IN
संबंधित प्रश्न
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (See the given figure). Prove that
- ΔABD ≅ ΔBAC
- BD = AC
- ∠ABD = ∠BAC.
You want to show that ΔART ≅ ΔPEN,
If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have
1) RT = and
2) PN =
In Fig. 10.92, it is given that AB = CD and AD = BC. Prove that ΔADC ≅ ΔCBA.
Which of the following statements are true (T) and which are false (F):
Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal equal to the hypotenuse and a side of the other triangle.
In the given figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is
D, E, F are the mid-point of the sides BC, CA and AB respectively of ΔABC. Then ΔDEF is congruent to triangle
The perpendicular bisectors of the sides of a triangle ABC meet at I.
Prove that: IA = IB = IC.
From the given diagram, in which ABCD is a parallelogram, ABL is a line segment and E is mid-point of BC.
Prove that:
(i) ΔDCE ≅ ΔLBE
(ii) AB = BL.
(iii) AL = 2DC
From the given diagram, in which ABCD is a parallelogram, ABL is a line segment and E is mid-point of BC.
Prove that: AB = BL.
In the adjoining figure, QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR.
If XS ⊥ QR and XT ⊥ PQ;
Prove that:
- ΔXTQ ≅ ΔXSQ.
- PX bisects angle P.
