Mathematics
Answer
Given: BC = CE and ∠1 = ∠2.
To prove: △GCB ≡ △DCE.
Proof: ∠1 + ∠GBC = 180°
∠2 + ∠DEC = 180°
From the above two equations,
⇒ ∠1 + ∠GBC = ∠2 + ∠DEC
∵ ∠1 = ∠2
Hence, ∠GBC = ∠DEC.
In △ GBC and △ DEC,
∠GBC = ∠DEC (Proved above)
BC = CE (Given)
∠GCB = ∠DCE (Vertically opposite angles)
By ASA congruency criteria,
△GCB ≅ △DCE
Hence, △GCB ≅ △DCE.
Related Questions
The given figure shows a right triangle right angled at B.
If ∠BCA = 2∠BAC, show that AC = 2BC.
In △ABC, AB = AC and D is a point in side BC such that AD bisects angle BAC.
Show that AD is perpendicular bisector of side BC.
The given figure shows two isosceles triangles ABC and DBC with common base BC. AD is extended to intersect BC at point P. Show that :
(i) △ABD ≡ △ACD.
(ii) △ABP ≡ △ACP.
(iii) AP bisects ∠BDC.
(iv) AP is perpendicular bisector of BC.
Two sides AB and BC and median AD of triangle ABC are respectively equal to sides PQ and QR and median PN of △PQR. Show that :
(i) △ABD ≡ △PQN.
(ii) △ABC ≡ △PQR.