In right triangle `A B C` , right angle at `C ,\ M` is the mid-point of the hypotenuse `A Bdot` `C` is jointed to `M` and produced to a point `D` such that `D M=C Mdot` Point `D` is joined to point `Bdot` Show that ` A M C~= B M D` (ii) `/_D B C=90^@` (iii)` D B C~= A CB` (iv) `C M=1/2A B`

Explanation

Given: `M` is the mid-point of hypotenuse `AB`, `∠C = 90°` and `DM = CM`
To Prove:
(i) `/_\AMC~=/_\BMD`
(ii) `/_DBC` is a right angle.
(iii) `/_\DBC~=/_\ACB`
(iv) `CM=1/2AB`
Proof:
(i) In `/_\AMC` and `/_\BMD`,
`AM=BM` (`M` is the mid - point of `AB`)
`/_AMC=/_BMD` (Vertically opposite angles)
`CM=DM` (Given)
`∴ /_\AMC~=/_\BMD` (By SAS congruence rule)
`∴ AC=BD` (By CPCT)
Also, `/_ACM=/_BDM` (By CPCT)
(ii) `/_DBC` is a right angle.
`/_ACM=/_BDM` (proved above)
But, `/_ACM` and `/_BDM` are alternate interior angles. Since alternate angles are equal, it can be said that `DB || AC`.
`/_DBC+/_ACB=180°` (Co-interior angles)
`/_DBC+90°=180°` [Since, `/_\ACB` is a right angled triangle]
`:./_\DBC = 90°`
Thus, `/_DBC` is a right angle.
(iii) In `/_\DBC` and `/_\ACB`,
`DB=AC` (proved above)
`/_DBC=/_ACB=90°` (Proved above)
`BC=CB` (Common)
`:./_\DB~=/_\ACB` (By `SAS` congruence rule)
(iv) `CM=1/2AB`
Since `/_\DBC~=/_\ACB`
`AB=DC` (By CPCT)
`=>1/2AB=1/2DC`

`CM=1/2DC=1/2AB`
`:. CM=1/2AB`
Hence proved.