In Triangle ABC, <BAC=90 Prove that the perpendicular bisectionrs of AB and AC will meet the midpoint of BC

Answer 1

In triangle ABC ,let EO be the perpendicular bisector of AB and PO be the perpendicular bisector of AC.BA is perpendicular AC and OP is perpendicular AC. therefore,EA is parallel to OP since the corresponding angles are equal.similarly OE parallel AP.
this implies ,EA=OP&EA=EB
therefore OP=EB -----------(1)
also,AP=EO&AP=PC
therefore EO=PC -----------(2)

in triangles BEO andOPC
BE=OP (from (1))
EO =PC(from(2))
therefore triangles BEO&OPC are congruent
i.e;BO=CO(CPCT)
i.e; Ois the midpoint of BC
i.e; the perpendicular bisectors ofAB & AC meet at the midpoint of BC

Answer 2

Let M be the mpt of AB and N be the mpt of AC. Perpendicular bisecors of AB and AC meet at O.
MONA is a rectangle (Prove it. Its easy.)
MO = AN = CN = n and NO = AM = BM = m
So we get triangles BMO, AMO, ONC and ONA congruent by SAS test.
Thus, we can prove angles BOM and AOM congruent and angles CON and AON congruent.
Now, angles BOM + MBO = 90 and angles MBO = CON = AON
Thus we can prove further that angles BMO, AMO, ONC and ONA add up to give 180.
So, B - O - C
BO = CO because these are c.s.c.t.
Thus, we have proved that O is the mpt of BA.
Now do it yourself. I can't draw figures here!
There is one more alternative method:
Let BA = 2m and CA = 2n
So, BCsquare = 4(m square + n square).....PGT
So, BC = 2 * squareroot(m square + n square)
Now, BM = MA = m and CN = NA = n
Triangle BMO is a rt. triangle
BO = squareroot(m square + n square)
Similarly, in triangle CNO,
CO = squareroot(m square + n square)
Thus, we get BO = CO and BO + CO = BC So, B - O - C
Thus, O is the mpt of BC

Answer 3

mid pt theorm

Answer 4

Use Pythagoras theorem and you will get the answer.

Answer 5

your question is wrong