In Triangle ABC, <BAC=90. Prove that the Perpendicular bisectors of AB and AC will meet the mid-point of BC.

Question

Let the Perpendicular bisectors of AB and AC meet at O. Then, O lies on the bisector of AB and AC. therefore OA=OB and OA=OC=>OB=OC=>O is the midpoint of BC.

Answer 1

If you approach this by drawing the perpendicular bisectors, you won't get very far, because you don't know that their intersection will be on the line BC.

Instead, you should join the midpoint of BC to the midpoints of AB and AC, and then try and prove that these two lines you have drawn are in fact the perpendicular bisectors.

One further line, joining midpoints of AB and AC divides your original triangle into four smaller triangles. If you now show that these four triangles are congruent, then it follows that the first two lines you drew are the perpendicular bisectors, and by construction they bisect BC.

Answer 2

Again, similar triangles come to the rescue. The perpendicular bisector of AC meets BC at some point D, and forms a triangle similar to the original triangle, but with each side half the length of the original. The same is true of the perpendicular bisector of AB. Since DB = DC, d must be the meeting point of both bisectors.

Answer 3

by using coordinate geometry also we can get it.

take A(0,0),B(0,b),C(c,0).where b,c>0.suppose that the perpendicular bisectors(pb) meet at O(c/2,b/2)

the pb of AC is x=c/2 & of AB is y=b/2.hence the result.

Answer 4

enable 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 (prepare it. Its uncomplicated.) MO = AN = CN = n and NO = AM = BM = m So we get triangles BMO, AMO, ONC and ONA congruent by applying SAS try. subsequently, we are able to coach angles BOM and AOM congruent and angles CON and AON congruent. Now, angles BOM + MBO = ninety and angles MBO = CON = AON subsequently we are able to coach extra that angles BMO, AMO, ONC and ONA upload as much as supply a hundred and eighty. So, B - O - C BO = CO via fact those are c.s.c.t. subsequently, we've proved that O is the mpt of BA. Now do it your self. i won't be able to draw figures right here! there is one greater option technique: enable BA = 2m and CA = 2n So, BCsquare = 4(m sq. + n sq.).....PGT So, BC = 2 * squareroot(m sq. + n sq.) Now, BM = MA = m and CN = NA = n Triangle BMO is a rt. triangle BO = squareroot(m sq. + n sq.) in addition, in triangle CNO, CO = squareroot(m sq. + n sq.) subsequently, we get BO = CO and BO + CO = BC So, B - O - C subsequently, O is the mpt of BC