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In right triangle ABC, right angle at C,...

In right triangle ABC, right angle at C, M is the mid-point of the hydrotenuse AB. C is joined to M and produced to a point D such that `DM=CM`. Point D is joined to point B. Show that
(i) `DeltaAMC cong DeltaBMD` (ii) `angleDBC=angleACB`
(iii) `DeltaDBC cong DeltaACB` (iv) `CM=1/2 AB`

Text Solution

Verified by Experts

(i) In `DeltaAMC` and `DeltaBMD`,
`:' {(AM=BM,( :'" M is the mid-point")),(angle1=angle2,"(vertically opposite angles)"),(CM=DM,"(given)"):}`
`:. DeltaAMC cong DeltaBMD` (SAS)
`implies" "AC=DB` (c.p.c.t) ...(1) Hence Proved.
(ii) `:. angle3=angle4` (c.p.c.t.)
But these are the alternate angles which are equal.
`:. AC||BD`
`:. (angle4+angle5)+angle6=180^(@)` (sum of co-int `angle5` angles)
`implies" "angle4+angle5=180^(@)-90^(@)" "( :' angle6=90^(@)", given")`
`implies" "angle4+angle5=90^(@)`
`implies" "angleDBC=90^(@)=angleACB` Hence Proved.
(iii) In `DeltaDBC` and `DeltaACB`,
`:' {(DB=AC,"[from (1)]"),(angleDBC=angleACB,"(just proved)"),(BC=BC,"(common)"):}`
`:. DeltaDBC cong DeltaACB` (ASA) Hence Proved.
(iv) `DC = AB` (c.p.c.t)
`implies" "2CM=AB` (`:'` M is the mid-point of DC)
`implies" "CM=1/2 AB`. Hence Proved.
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