In the given figure, ABC is a triangle right-angled at B such that ∠BCA=2∠BAC.
Show that AC=2BC.
We will have to make the following construction in the given figure:
Produce CB to D in such a way that BD=BC and join AD.
Now, in ∆ABC and ∆ABD,
BC=BD (constructed)
AB=AB (common)
∠ABC=∠ABD (each 90°)
∴ by S.A.S.
∆ABC ≅ ∆ABD
∠CAB=∠DAB and AC=AD (by c.p.c.t.)
∴∠CAD=∠CAB+∠BAD
=x°+x°
=2x°
But, AC=AD
∠ACD=∠ADB=2x°
∴ ∆ACD is equilateral triangle.
AC=CD
AC=2BC
Hence, proved