Question

In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are
squares on the sides BC, CA and AB respectively. Line segment AXI DE meets BC
at Y. Show that:
G​

Answer 1

Answer:

Since all angle of square is 90 degree

angle ABM=90

angle DBC=90

Now

<ABM=<DBC

<ABM+<ABC=<OBC+<ABC

<MBC=<ABD------1

In ∆MBC and ∆ABD,

MB=AB. [Same sides of a square are equal]

<MBC=<ABD [From 1]

BC=BD [Same sides of a square are equal]

∆MBC ≈∆ABD. [SAS]